The ratio of cube of circumference of the orbit of a satellite to the volume of the earth is 6×1010(gR), where g is the acceleration of gravity. Find the time period of satellite (in seconds). R is the radius of earth.

T = 2πrRrg = KRK2πg If K = 2πr = 2K33R(43πr3)G = 2K3R3 VgGiven, K3V = 6×1010RSo, T = 2×6×10103 = 2×105 sec

The ratio of cube of circumference of the orbit of a satellite to the volume of the earth is 6×1010(gR), where g is the acceleration of gravity. Find the time period of satellite (in seconds). R is the radius of earth.

T = 2πrRrg = KRK2πg If K = 2πr = 2K33R(43πr3)G = 2K3R3 VgGiven, K3V = 6×1010RSo, T = 2×6×10103 = 2×105 sec

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The ratio of cube of circumference of the orbit of a satellite to the volume of the earth is $6 \times 10^{10} (\frac{g}{R})$ , where g is the acceleration of gravity. Find the time period of satellite (in seconds). R is the radius of earth.

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$2 \times 10^{5}$

$10^{5}$

$4 \times 10^{4}$

$10^{4}$

answer is A.

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T = $\frac{2 πr}{R} \sqrt{\frac{r}{g}} = \frac{K}{R} \sqrt{\frac{K}{2 πg}} If K = 2 πr$

= $\sqrt{\frac{\frac{2 K^{3}}{3} R}{(\frac{4}{3} {πr}^{3}) G}} = \sqrt{\frac{2 K^{3} R}{3 Vg}}$

Given, $\frac{K^{3}}{V} = \frac{6 \times 10^{10}}{R}$

So, $T = \sqrt{\frac{2 \times 6 \times 10^{10}}{3}} = 2 \times 10^{5} \sec$

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The ratio of cube of circumference of the orbit of a satellite to the volume of the earth is 6×1010(gR), where g is the acceleration of gravity. Find the time period of satellite (in seconds). R is the radius of earth.

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The ratio of cube of circumference of the orbit of a satellite to the volume of the earth is $6 \times 10^{10} (\frac{g}{R})$ , where g is the acceleration of gravity. Find the time period of satellite (in seconds). R is the radius of earth.